Why did I get an irreducible factor left over?

That means none of the remaining Rational Root Theorem candidates worked on what was left. The leftover piece still has roots, they are just irrational or complex rather than rational, which this method cannot isolate individually.

Roots of a Polynomial Calculator

Automatically find every rational root of a polynomial — no manual candidate testing required.

x³−6x²+11x−6Three Roots

x⁴−1Irreducible Remainder

2x³−3x²−11x+6Fractional Root

x³−3x²+4Repeated Root

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Roots

Answer

Steps go here…

What This Calculator Automates

Finding roots by hand usually means three separate tools working together: the Rational Root Theorem to generate a list of candidates, the Factor Theorem to test each one, and synthetic division to divide out each confirmed root and shrink the polynomial before searching again.

This calculator runs that whole loop for you — generate candidates, test, divide, repeat — until it either factors the polynomial completely or is left with a piece that has no more rational roots.

The search loop

generate candidates \to test with P(c) \to divide out confirmed roots \to repeat

Two Worked Examples

Example 1 — a fully factorable cubic

Find all roots of: x³ − 6x² + 11x − 6

Step 1. Constant term is −6, leading coefficient is 1. Candidates: ±1, ±2, ±3, ±6.

Step 2. Testing finds P(1) = 0, so x = 1 is a root. Dividing it out leaves x² − 5x + 6.

Step 3. That quadratic factors directly (or keep testing candidates): roots x = 2 and x = 3.

Final answer: x = 1, 2, 3 — all three roots are rational, and the polynomial factors completely as (x − 1)(x − 2)(x − 3).

Example 2 — a leftover that won’t factor further

Find all roots of: x⁴ − 1

Step 1. Candidates from ±1 (the only divisor of both 1 and −1): x = 1 and x = −1.

Step 2. Both check out: P(1) = 0 and P(−1) = 0. Dividing both out leaves x² + 1.

Step 3. Testing ±1 again on x² + 1 fails — there’s no rational root here because x² + 1 = 0 only has the complex solutions x = ±i.

Final answer: x = 1, −1, with x² + 1 left over as an irreducible remainder over the rational numbers.

Common Mistakes

Assuming every polynomial has all-rational roots. Many don’t — a leftover quadratic or higher-degree piece with no rational root is a completely normal, correct outcome.
Forgetting repeated roots. A root can show up more than once (like x = 2 appearing twice in x³ − 3x² + 4) — after dividing it out the first time, test the same candidate again on what’s left.
Stopping the candidate list too early. Every divisor of the constant term over every divisor of the leading coefficient needs to be checked, including the negative versions of each.
Missing fractional candidates. If the leading coefficient isn’t 1, roots like 1/2 or 3/4 become possible and are easy to overlook if you only test whole numbers.

Frequently Asked Questions

No, only rational roots. If the polynomial has irrational roots (like √2) or complex roots (involving i), those show up as an irreducible remaining factor instead of an individual root — you’d then reach for the Quadratic Formula Calculator if that leftover piece is degree 2.

The Rational Root Theorem generates candidates as plus-or-minus a ratio of divisors, since a root can be negative even when both the constant term and leading coefficient are positive.

It means that root appears more than once as a factor. A root with multiplicity 2, like x = 2 in x^3 – 3x^2 + 4, means (x – 2) appears twice in the fully factored form.

That means none of the remaining Rational Root Theorem candidates worked on what was left. The leftover piece still has roots, they’re just irrational or complex rather than rational, which this method can’t isolate individually.